Comparing a 2D array with a number

Well, you try to use list's .all() method, which does not exist, so your code shouldn't work at all:

>>> a = [1, 2, 3]
>>> a.all()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'all'

And if it were a NumPy array, as per @AKX's comment, its .all(…) method would just test if all elements evaluate to True, but negative integers also evaluate to True in Python:

>>> bool(-1)
True

What you should do is use built-in all(…) function:

a = [[1, 2, 3, 4, 5], [1, 2, 3, 4, 6], [-1, 2, 3, 4, -5]]
d = 0
for inner_list in a:
    if all(x > 0 for x in inner_list):
        d += 1

Or you could use sum(…) function:

d = sum(1 for inner_list in a if all(x > 0 for x in inner_list))

As you can see, I've used for inner_list in a in both cases, which is just a nicer way to iterate over list's elements than using its indexes — in your case it would actually be equivalent to:

for i in range(len(a)):  # or "in range(3)" in your specific case
    inner_list = a[x]
    ...

a = [[1,2,3,4,5],[1,2,3,4,6],[-1,2,3,4,-5]]
d = 0
for x in range(3):
    if (all(e > 0 for e in a[x])):
        d = d+1

For a[x] equal to [-1,2,3,4,-5], [e > 0 for e in a[x]] is equal to [False, True, True, True, False]. Then, all(e > 0 for e in a[x]) checks if all values in [False, True, True, True, False] are True.

below is the code which might help

a=np.array([[1,2,3,4,5],[1,2,3,4,6],[-1,2,3,4,-5]])
    d=0
    for iter1 in range (a.shape[0]):
        flag = True
        for iter2 in range (a.shape[1]):
            print (a[iter1][iter2])
            if a[iter1][iter2] < 0:
                flag = False

        print("moving to next row check the flag value="+ str(flag))
        if flag:
            d = d+1
    print("Number of profiles which are positive="+ str(d))